# 009A Sample Final A, Problem 2

2. Find the derivatives of the following functions:
(a)  ${\displaystyle f(x)={\frac {3x^{2}-5}{x^{3}-9}}.}$

(b)  ${\displaystyle g(x)=\pi +2\cos({\sqrt {x-2}}).}$

(c)
${\displaystyle h(x)=4x\sin(x)+e(x^{2}+2)^{2}.}$

Foundations:
These are problems involving some more advanced rules of differentiation. In particular, they use
The Chain Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$

The Product Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).}$

The Quotient Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions and ${\displaystyle g(x)\neq 0}$ , then

${\displaystyle \left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.}$

Solution:

Part (a):
We need to use the quotient rule:
${\displaystyle f'(x)={\frac {\left(3x^{2}-5\right)'\cdot \left(x^{3}-9\right)-\left(3x^{2}-5\right)\cdot \left(x^{3}-9\right)'}{\left(x^{3}-9\right)^{2}}}}$
${\displaystyle ={\frac {6x(x^{3}-9)-(3x^{2}-5)(3x^{2})}{(x^{3}-9)^{2}}}}$
${\displaystyle ={\frac {6x^{4}-54x-9x^{4}+15x^{2}}{(x^{3}-9)^{2}}}}$
${\displaystyle ={\frac {-3x^{4}+15x^{2}-54x}{(x^{3}-9)^{2}}}.}$
Part (b):
Both parts (b) and (c) attempt to confuse you by including the familiar constants ${\displaystyle e}$ and ${\displaystyle \pi }$. Remember - they are just constants, like 10 or 1/2. With that in mind, we really just need to apply the chain rule to find
${\displaystyle g'(x)\,=\,0-2\sin \left({\sqrt {x-2}}\right)\cdot {\frac {1}{2}}\cdot (x-2)^{-1/2}\cdot 1\,=\,\,-\,{\frac {\sin \left({\sqrt {x-2}}\right)}{\sqrt {x-2}}}.}$
Part (c):
We can choose to expand the second term, finding
${\displaystyle e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.}$
We then only require the product rule on the first term, so
${\displaystyle h'(x)\,=\,(4x)'\cdot \sin(x)+4x\cdot (\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.}$